$A=\left[\begin{array}{rr}-28 & 17 & -13 & 15 & 29 \\-3 & 17 & 21 & -17 & 9 \\3 &-3 & 2 & 8 & 1 \\1 &1 &2 & 6 & 5 \\6 &5 &9 & 2 & 1\end{array}\right]$ $A_{1,5}=$
Answer: Background An $m\times n$ matrix has $m$ rows and $n$ columns. $A=\left[\begin{array}{rr}A_{1,1} & \cdots & A_{1,n} \\\\\vdots \ & \ddots & \vdots \\\\A_{m,1} &\cdots &A_{m,n}\end{array}\right]$ Therefore, the entry $A_{{c},{d}}$ is located on row ${c}$ and column ${d}$. Finding $A_{1,5}$ $A_{{1},{5}}$ is located on row ${1}$ of $A$ : $\left[\begin{array}{rr}{-28} & {17} & {-13} & {15} & {29} \\-3 & 17 & 21 & -17 & 9 \\3 &-3 & 2 & 8 & 1 \\1 &1 &2 & 6 & 5 \\6 &5 &9 & 2 & 1\end{array}\right]$ $A_{{1},{5}}$ is also located on column ${5}$ of $A$. $\left[\begin{array}{rr}{-28} & {17} & {-13} & {15} & {\text{29}} \\-3 & 17 & 21 & -17 & {9} \\3 &-3 & 2 & 8 & {1} \\1 &1 &2 & 6 & {5} \\6 &5 &9 & 2 & {1}\end{array}\right]$ Therefore, $A_{{1},{5}}={29}$. Summary $A_{1,5}=29$